Degenerate Linear Operator Plus Identity is Fredholm Operator
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Theorem
Let $U$ be a vector space.
Let $T : U \to U$ be a degenerate linear operator.
Let $I_U : U \to U$ be the identity operator.
Then:
- $T + I_U$
is a Fredholm operator.
Proof
We need to show that both:
- $\map \ker {T + I_U}$
and:
- $U / {\Img {T + I_U} }$
are finite-dimensional.
Recall that:
- $(1):\quad \map \dim {\Img T} < +\infty$
since $T$ is degenerate.
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$\map \dim {\map \ker {T + I_U} } < +\infty$
Let $u \in \map \ker {T + I_U}$.
That is:
- $Tu + u = 0$
Thus:
- $u = \map T {-u} \in \Img T$
Hence:
- $\map \ker {T + I_U} \subseteq \Img T$
Therefore:
\(\ds \map \dim {\map \ker {T + I_U} }\) | \(\le\) | \(\ds \map \dim {\Img T}\) | Dimension of Proper Subspace is Less Than its Superspace | |||||||||||
\(\ds \) | \(<\) | \(\ds + \infty\) | by $(1)$ |
$\Box$
$\map \dim {U / {\Img {T + I_U} } } < +\infty$
Observe that:
- $\map \ker T \subseteq \Img {T + I_U}$
since if $u \in \map \ker T$, then:
- $u = T u + u \in \Img {T + I_U}$
Thus:
\(\ds \map \dim {U / {\Img {T + I_U} } }\) | \(=\) | \(\ds \map {\mathrm {codim} } {\Img {T + I_U} }\) | Definition of Codimension of Vector Subspace | |||||||||||
\(\ds \) | \(\le\) | \(\ds \map {\mathrm {codim} } {\map \ker T}\) | Codimension of Proper Subspace is Greater | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \dim {U / \map \ker T}\) | Definition of Codimension of Vector Subspace | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \dim {\Img T}\) | $U / \map \ker T \cong \Img T$ by First Isomorphism Theorem | |||||||||||
\(\ds \) | \(<\) | \(\ds + \infty\) | by $(1)$ |
$\blacksquare$
Sources
- 2002: Peter D. Lax: Functional Analysis: Chapter $27$: Index Theory